91Ó°ÊÓ

Consider the given question,

The mean mileage is 21.4 mpg and standard deviation is 1.15 mpg.

On assuming the advertised claim correct which is buying a new car whose advertised mileage is 25 miles per gallon (mpg).

The z-score for the gas mileage of your car,

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{21.4-25}{1.15} \\ =-3.1304 \\ ≈-3.13 \end{gathered}$$

For unusual mileage we have to find the one, two and three standard deviation of the mean.

Unusual observations lie between the ranges $\overline{x}Â\pm \mathrm{σ}$,

$$\begin{gathered} =\left(\overline{x}-\mathrm{σ},\overline{x}+\mathrm{σ}\right) \\ =\left(25-1.15,25+1.15\right) \\ =\left(23.85,26.15\right) \end{gathered}$$

Unusual observations lie between the ranges $\overline{x}Â\pm 2\mathrm{σ}$,

$$\begin{gathered} =\left(\overline{x}-2\mathrm{σ},\overline{x}+2\mathrm{σ}\right) \\ =\left(25-2×1.15,25+2×1.15\right) \\ =\left(21.55,28.45\right) \end{gathered}$$

By observing the three standard deviation rule implies that your car's mileage is lower than most other cars of the model.

Hence, our assumption the advertised claim is correct.